What is the Potential Energy of a spring stretched 5.0 cm that has a
spring constant of 79 N/m?

The Potential Energy of the stretched spring is = 1.975 J

Explanation:

Recall that the potential energy of a spring of constant k stretched x\a distance x is given by the formula:

U = 1/2  k x^2

Therefore in our case, converting the stretch from cm of meters (0.05 m), this is

U = 1/2 * 79 * 0.05 J = 1.975 J

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.

it is given that trillions of microbes make up to 1% to 3% of the body mass.

let's take average of this percentage: 2%

consider the average mass of the human body as 70 kg.

then, trillions of microbes would have mass  

therefore, 100 trillion micro-organisms would weigh:

this is an approximate value and could change depending on the mass of the human body and percentage of the microbes present in it.

average mass per microorganism =